Please Help! :D Polynomial Question (1 Viewer)

[mawissah]

The polynomial f(x) is given by f(x) = px³ + 16x² + qx - 120, where p and q are constants. The three zeroes of f(X) are (-2), 3 and a. The value of a is?

I'm not sure how to start. and what does 'zeroes' mean? How do work this out?

It was a multiple choice so the answer is either, 5/9, 4, (-5) or 5

 

[Capt Rifle]

when you sub in f(-2), f(3) and f(a) it should equal 0. because (x+2) and (x-3) and (x-a) are factors of that polynomial equation

 

[Capt Rifle]

so solve for a in that manner

 

[mawissah]

Alternatively,

Zeroes = roots = -2, 3, a

= Sum of roots = -2 + 3 + a = -16

woah, where did the -16 come from?

 

[HeroicPandas]

woah, where did the -16 come from?

Sorry nvm, honest mistake

 

[hayabusaboston]

Huuh isnt sum of roots the way to go here? meaning a is -17? i dont see that in the MC answers?

 

[wheel]

The polynomial f(x) is given by f(x) = px³ + 16x² + qx - 120, where p and q are constants. The three zeroes of f(X) are (-2), 3 and a. The value of a is?

I'm not sure how to start. and what does 'zeroes' mean? How do work this out?

It was a multiple choice so the answer is either, 5/9, 4, (-5) or 5

zeros is another word for roots or solutions.

by the factor theorem, if you substitute in (-2), (3) or (a), P(x) = 0

so sub in (-2) and (3) for x and solve it for P(x) = 0

by simultaneous equations you will find p and q

then sub in a for x with the results of p and q in the polynomial and solve P(a) = 0

goodluck and i hope this helps.

 

[lochnessmonsta]

This can be solved much simpler by sum and product of roots

ie sum of roots ---> -2 + 3 + a = -16/p
product of roots ---> -6a = 120/p
p = -20/a

sub p into the first equation

1 + a = -16/(-20/a)
1 + a = 4a/5
1 = -a/5
a = -5

 

[lochnessmonsta]

inb4 not monic polynomial

 

[hayabusaboston]

woah lol, well that was stupid of me haha, not to notice non monic thing. Just like heroicpandas LOL

 

[HeroicPandas]

woah lol, well that was stupid of me haha, not to notice non monic thing. Just like heroicpandas LOL

lol

 

[hayabusaboston]

K gaiz! Im feelin motivated! someone fire up a 'nomials question 'ere!

 

[HeroicPandas]

K gaiz! Im feelin motivated! someone fire up a 'nomials question 'ere!

 

[hayabusaboston]

Sorry took so long brah, I couldnt get ur result, I think u made mistake? here's my process:

Since alpha is root, divide x^3+lx+b by (x-alpha). Remainder is alpha^3+l.alpha+b, which is equal to zero anyway, since it satisfies equation of 'nomial, and result is (x^2+alpha.x+(l+alpha^2)) Rewriting the polynomial, you get the product of division times by (x-alpha), ie

(x-alpha)(x^2+alpha.x+(l+alpha^2))

for real roots, discriminant >= zero, ie
alpha^2-4(l+alpha^2)>=0
alpha^2-4l-4alpha^2>=0
-3alpha^2-4l>=0

ie 4l+3alpha^2>=0

Did you forget the ^2 on the 3alpha?

 

[HeroicPandas]

Sorry took so long brah, I couldnt get ur result, I think u made mistake? here's my process:

Since alpha is root, divide x^3+lx+b by (x-alpha). Remainder is alpha^3+l.alpha+b, which is equal to zero anyway, since it satisfies equation of 'nomial, and result is (x^2+alpha.x+(l+alpha^2)) Rewriting the polynomial, you get the product of division times by (x-alpha), ie

(x-alpha)(x^2+alpha.x+(l+alpha^2))

for real roots, discriminant >= zero, ie
alpha^2-4(l+alpha^2)>=0
alpha^2-4l-4alpha^2>=0
-3alpha^2-4l>=0

ie 4l+3alpha^2>=0

Did you forget the ^2 on the 3alpha?

Yes i did haha! Good job

 

[HeroicPandas]

Sorry took so long brah, I couldnt get ur result, I think u made mistake? here's my process:

Since alpha is root, divide x^3+lx+b by (x-alpha). Remainder is alpha^3+l.alpha+b, which is equal to zero anyway, since it satisfies equation of 'nomial, and result is (x^2+alpha.x+(l+alpha^2)) Rewriting the polynomial, you get the product of division times by (x-alpha), ie

(x-alpha)(x^2+alpha.x+(l+alpha^2))

for real roots, discriminant >= zero, ie
alpha^2-4(l+alpha^2)>=0
alpha^2-4l-4alpha^2>=0
-3alpha^2-4l>=0

ie 4l+3alpha^2>=0

Did you forget the ^2 on the 3alpha?

Yes i did haha! Good job

 

[hayabusaboston]

Is Sy123 aRound the place? I wanna show him I can do his questions haha.